序列化不是为PgRange<DateTime<Utc>>实现的
序列化不是为PgRange<DateTime<Utc>>实现的
提问于 2022-07-06 07:38:46
但是，我希望能够序列化包含PgRange<DateTime<Utc>>的结构，但是，#[derive(Serialize)]对struct Reservation的错误失败：
pub timespan: PgRange<DateTime<Utc>>,
^^^ the trait `Serialize` is not implemented for `PgRange<chrono::DateTime<chrono::Utc>>`作为一种解决方法，我将timespan字段分成两个字段，用于序列化，但使SQL更加复杂。
数据方案如下：
CREATE TABLE reservation (
id INTEGER PRIMARY KEY GENERATED BY DEFAULT AS IDENTITY,
timespan TSTZRANGE
);
INSERT INTO reservation(timespan) VALUES
(TSTZRANGE(now() + INTERVAL '0 hour', now() + INTERVAL '1 hour')),
(TSTZRANGE(now() + INTERVAL '2 hour', now() + INTERVAL '3 hour'));还有我的工作代码示例：
/*
[dependencies]
chrono = { version = "0.4", features = ["serde"] }
tokio = { version = "1", features = ["full"] }
sqlx = { version = "0.5", features = ["runtime-tokio-native-tls" , "postgres", "chrono"] }
serde = { version = "1.0", features = ["derive"] }
serde_json = "1.0"
*/
use chrono::prelude::*;
use serde::{Serialize, Deserialize};
use sqlx::FromRow;
use sqlx::postgres::types::PgRange;
use sqlx::postgres::PgPoolOptions;
#[derive(FromRow, Debug)]
pub struct Reservation {
pub id: i32,
pub timespan: PgRange<DateTime<Utc>>,
}
#[derive(FromRow, Serialize, Debug)]
pub struct ReservationWorkaroud {
pub id: i32,
pub start: DateTime<Utc>,
pub end: DateTime<Utc>,
}
#[tokio::main]
async fn main() -> Result<(), sqlx::Error> {
let pool = PgPoolOptions::new()
.max_connections(2)
.connect(&"postgresql://postgres:postgres@localhost")
.await?;
// Ideally, I would like my struct to mirror db table as close as possible while being able to serialize/deserialize to/from Json
let select_query = sqlx::query_as::<_, Reservation>("SELECT id, timespan FROM reservation");
let reservations: Vec<Reservation> = select_query.fetch_all(&pool).await?;
dbg!("{:?}", reservations);
// With the workaroud struct, I am able to serialize it however, sql queries start to get complex especially for inserting:
let select_query = sqlx::query_as::<_, ReservationWorkaroud>("SELECT id, lower(timespan) AS start, upper(timespan) AS end FROM reservation");
let reservations: Vec<ReservationWorkaroud> = select_query.fetch_all(&pool).await?;
dbg!("{:?}", reservations);
Ok(())
}如何使struct Reservation可序列化？
更新
基于@恺撒的回答，我采用了以下方法：
#[derive(FromRow, Serialize,Debug)]
pub struct Reservation {
pub id: i32,
#[serde(serialize_with = "serialize_range", flatten)]
pub timespan: PgRange<DateTime<Utc>>,
}
fn serialize_range<S, T>(range: &PgRange<T>, serializer: S) -> Result<S::Ok, S::Error>
where
S: serde::Serializer,
T: Serialize,
{
let PgRange { start, end } = range;
std::ops::Range { start, end }.serialize(serializer)
}回答 1
Stack Overflow用户
回答已采纳
发布于 2022-07-07 01:54:45
关于这一点，有很多种方法可以继续下去。
- 我最喜欢的serde技巧，使用容器属性
#[serde(into = "ReservationWorkaround")]onstruct Reservation并添加一个impl Into<ReservationWorkaround> for Reservation {}。有关使用属性的示例，请参见这里。 - 使用自定义序列化函数
#[derive(FromRow, Debug, Serialize)]
pub struct Reservation {
pub id: i32,
#[serde(serialize_with = "serialize_range", flatten)]
pub timespan: PgRange<DateTime<Utc>>,
}
fn serialize_range<S, T>(range: &PgRange<T>, serializer: S) -> Result<S::Ok, S::Error>
where
S: serde::Serializer,
T: Serialize,
{
// You probably also want to convert start and end
// from Bound<DateTime<Utc>> to something else here.
// It is common to define a new struct like
#[derive(Serialize)] struct I64Range { start: i64, end: i64 }
// right here in this function
let PgRange { start, end } = range;
std::ops::Range { start, end }.serialize(serializer)
}- 理论上，这是“远程派生”的主要用例，但是您必须想出一个单独的解决方案来转换
Bound<DateTime<Utc>>。也许serde_with箱子能帮上忙，但我不确定。
#[derive(FromRow, Debug, Serialize)]
pub struct Reservation {
pub id: i32,
#[serde(with = "PgRangeRemote", flatten)]
pub timespan: PgRange<DateTime<Utc>>,
}
#[derive(Serialize)]
#[serde(remote = "PgRange")]
struct PgRangeRemote<T> {
start: std::ops::Bound<T>,
end: std::ops::Bound<T>,
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/72879584
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